Max Dehn 1912 Transformation of curves on two-sided surfaces Transformation der Kurven auf zweiseitigen Flftchen. by Von M. Dehn in Kiel M. D~H~ in Kie]. The problem we shall consider in what follows is one of the simplest in topology: Given two closed curves on a closed two-sided surface, Das Problem, das uns im fblgenden beseh~iftigen wird, ist eines der decide whether whether they can be “transformed into each other” einfachsten der Topologie: Gegeben xlnd zwei geschlossene Kurven auf dner by continuous deformation. The solution of the problem for surfaces 9eschlossenen zweiseitigen Fl~he~ es ist zu untersuchen, ob sie d~rch s~tige with genus g >1, with the help of “polygon groups” and accordingly Deformation ineinander i~bergefiihrt, ,,ineinander transformierU werden k6nnen. Die LSsung des Problems ffir Fl~ichen mit einem Geschlecht p > 1 mit based on the metric of the hyperbolic plane, is obvious .... Hilfe der ,Polygongruppen" und demgem~B auf Grund tier Metrik der hyperbolischen Ebene ist naheliegend mid z. B. yon Poincar6 (Rend. Circ~ Mat. Pal. 1905) angedeutet, yon mir in der A_rbeit Math. Ann. 71 ganz genau en~ickelt.*) In derselben Arbeit babe ieh auch eine Methode angegeben, um ohne Hilfe der Metrik rein topologisch die Frage zu entscheiden. Bei der Begriindun9 dieser Methode babe ieh aber sehr wesenflich Eigen- scha~n yon Figuren tier hyperbolischen Ebene benutzt. -- Fiir Flgehen yore Geschlecht p = 0 und p = 1 ist die LSsung des Problems sehr ein- fach: im ersten Fall sind alle Kurven ineinander transformierbar, im zweiten Fall ist die ,,Fundamentalgruppe" abelsch, und jede Kurve ist trans-
Ancient history ‣ Solvable via covering spaces [Schwarz 1870s, Poincaré 1905] ‣ Algorithms using hyperbolic geometry [Dehn 1911] ‣ Dehn 1912: Purely symbolic/combinatorial algorithms ▹ Dehn’s technique now called small cancellation [Lyndon Schupp 1977, Epstein et al 1992, McCammond Wise 2000] ▹ For any fixed surface, Dehn’s algorithm runs in O(ℓ) time, where ℓ = length of input curve(s).
Input ‣ Surface map Σ with complexity n and genus g ‣ A closed walk of length ℓ in the graph of Σ
Input This is what “given a surface” means! ‣ Surface map Σ with complexity n and genus g ‣ A closed walk of length ℓ in the graph of Σ
Input This is what “given a surface” means! ‣ Surface map Σ with complexity n and genus g ‣ A closed walk of length ℓ in the graph of Σ This is what “given a curve” means!
Dehn’s algorithm
System of loops ‣ 2g loops with common basepoint, which cut the surface into a disk ‣ Every cycle in Σ is homotopic to a concatenation of these loops. [Jordan 1866] ‣ Dehn assumes that the input cycle is a walk in this graph.
System of loops ‣ Contract edges between distinct vertices (spanning tree) ‣ Delete edges separating distinct faces (spanning cotree) ‣ Euler’s formula ⇒ exactly 2 g edges left
System of loops ‣ Cut the surface along 2g loops → fundamental polygon P with 4 g edges ‣ Edge in G → path of length ≤2 g around boundary of P ‣ Walk of length ℓ in G → walk of length ℓ ʹ ≤2 gℓ in loop system b b b a a a a a a d c d c d c c c c d d d b b b
Universal cover Σ̃ ‣ Cut surface along loops → fundamental polygon ‣ Glue infinitely many fundamental polygons into a plane.
Universal cover Σ̃ When g >1, the universal cover is a regular tiling of the hyperbolic plane by 4 g -gons with vertices of degree 4 g. b a a d c c d b
Universal cover Σ̃ ‣ Any point in Σ has infinitely many lifts in the universal cover, one for each homotopy class of loops. ‣ Every path has infinitely many lifts, each determined by a lift of one endpoint. Just follow the “same” sequence of edges. ‣ A loop in Σ is contractible iff it lifts to a loop in Σ̃.
Universal cover Σ̃ ‣ Any point in Σ has infinitely many lifts in the universal cover, one for each homotopy class of loops. ‣ Every path has infinitely many lifts, each determined by a lift of one endpoint. Just follow the “same” sequence of edges. ‣ A loop in Σ is contractible iff it lifts to a loop in Σ̃.
Universal cover Σ̃ ‣ Any point in Σ has infinitely many lifts in the universal cover, one for each homotopy class of loops. ‣ Every path has infinitely many lifts, each determined by a lift of one endpoint. Just follow the “same” sequence of edges. ‣ A loop in Σ is contractible iff it lifts to a loop in Σ̃.
Universal cover Σ̃ ‣ Any point in Σ has infinitely many lifts in the universal cover, one for each homotopy class of loops. ‣ Every path has infinitely many lifts, each determined by a lift of one endpoint. Just follow the “same” sequence of edges. ‣ A loop in Σ is contractible iff it lifts to a loop in Σ̃.
Dehn’s algorithm [Dehn 1912] ‣ Goal: Find a shortest cycle homotopic to the given cycle α . ‣ Algorithm: Greedily reduce α by local curve shortening. ‣ Hyperbolic structure ⇒ no local minima! ‣ Thus α is contractible iff it reduces to nothing.
Dehn’s algorithm [Dehn 1912] ‣ Key Lemma: Every nontrivial closed walk in Σ̃ contains either a spur or 4 g –2 consecutive edges of some face. ‣ Thus, we can locally shorten any contractible cycle.
Dehn’s algorithm [Dehn 1912] ‣ Key Lemma: Every nontrivial closed walk in Σ̃ contains either a spur or 4 g –2 consecutive edges of some face. ‣ Thus, we can locally shorten any contractible cycle.
René Descartes (c.1630) DE SouooRuM ELEMENTIS. 2t8-2t9. Progymnasmata De Solidorum Elementis Angulorum folidorum inclinatione requaliumjhac capacitate major eft, qui arithmetice exuperat ; & omnium capaciffimus eft angulus coni. Ponam femper pro numero angulorum folidorum C1. s & pro numero facierum <fl· Aggregatum ex omnibus angulis planis eft 4 a.- 8, & numerus cp eft 2 a.- 4, fi numerentur tot facies quot polfunt elfe triangula. Numerus item angulorum planorum eft 6 (1. -1 2, nume- rando fcilicet vnum angulum pro tertia parte duorum 10 retl:orum. Nunc fi. ponam J (1. pro tribus angulis planis qui ad minimum requiruntur vt componant vnum angulum angulorum folidorum, fuperfunt J (1.- 12, qure fumma addi debet fingulis angulis folidis juxta tenorem qureftionis, ita vt requaliter omni ex parte .s diffundantur. Numerus verorum angulorum planorum eft 2 <p + 2 C1.- 4, qui non debet elfe major quam 6 C1. 12; fed fi minor eft, excelfus erit + 4 (1.- 8- 2 <fl· - Defcribi polfunt & rhomboides in cujuf- cumque quantitatis, fed non requilaterre. (II) 20 Omnium a optime formabuntur folida per gnomones fuperadditos vno femper angulo vacuo exiftente, ac deinde totam figuram refolvi polfe in triangula. Vnde facile agnofcitur omnium polygonalium pondera haberi . ex multipli!catione trigonalium per numeros z5 - 2, 3, 4, ) , 6, &c., & ex produtl:o fi toll'a.ntur 1, 2, .3, 4, radices, &c. a. Dans le MS. aucune separation n'existe entre ce nouveau develop- pement et celui qui precede. Nous ajoutons (II), comme (I), p. 26S. .
René Descartes (c.1630) DE SouooRuM ELEMENTIS. 2t8-2t9. Progymnasmata De Solidorum Elementis Angulorum folidorum inclinatione requaliumjhac capacitate major eft, qui arithmetice exuperat ; & Let α always denote the number of solid angles and φ omnium capaciffimus eft angulus coni. the number of faces. The total of all plane angles is Ponam femper pro numero angulorum folidorum C1. s & pro numero facierum <fl· Aggregatum ex omnibus 4α–8 [right angles], and the number φ is 2α–4, if as angulis planis eft 4 a.- 8, & numerus cp eft 2 a.- 4, fi many faces as possible are triangles. The number of numerentur tot facies quot polfunt elfe triangula. planar angles themselves is 6α–12, counting for each Numerus item angulorum planorum eft 6 (1. -1 2, nume- angle a third part of two right angles. Then if I take 3α rando fcilicet vnum angulum pro tertia parte duorum for the three planar angles that are required at 10 retl:orum. Nunc fi. ponam J (1. pro tribus angulis planis minimum to comprise one angle of a solid angle, there qui ad minimum requiruntur vt componant vnum remain 3α–12 that must be added to the solid angles, angulum angulorum folidorum, fuperfunt J (1.- 12, qure fumma addi debet fingulis angulis folidis juxta according to the terms of the question, so that they are tenorem qureftionis, ita vt requaliter omni ex parte distributed equally to all parts. The total number of .s diffundantur. Numerus verorum angulorum planorum plane angles is 2φ–2α–4 , which cannot be larger than eft 2 <p + 2 C1.- 4, qui non debet elfe major quam 6 C1. 6α–12; if it is less, the excess is 4α–8–2φ. 12; fed fi minor eft, excelfus erit + 4 (1.- 8- 2 <fl· - Defcribi polfunt & rhomboides in cujuf- cumque quantitatis, fed non requilaterre. (II) 20 Omnium a optime formabuntur folida per gnomones fuperadditos vno femper angulo vacuo exiftente, ac deinde totam figuram refolvi polfe in triangula. Vnde facile agnofcitur omnium polygonalium pondera haberi . ex multipli!catione trigonalium per numeros z5 - 2, 3, 4, ) , 6, &c., & ex produtl:o fi toll'a.ntur 1, 2, .3, 4, radices, &c. a. Dans le MS. aucune separation n'existe entre ce nouveau develop- pement et celui qui precede. Nous ajoutons (II), comme (I), p. 26S. .
Combinatorial Gauss-Bonnet Theorem ‣ Consider any surface map ( V , E , F ), possibly with boundary ‣ Assign an arbitrary exterior angle ∠ c to every corner c κ ( v ) := 1 − 1 ‣ Vertex curvature : 2 deg( v ) + ∑ c ∈ v ∠ c ‣ Face curvature : κ ( f ) := 1 − ∑ c ∈ f ∠ c ∑ v κ ( v ) + ∑ f κ ( f ) = χ = V − E + F [Descartes 1630, Hilbert Cohn-Vossen 1932, Pólya 1954, Lyndon 1966, Banchoff 1967, Gromov 1987, McCammond Wise 2000, ...]
Dehn’s lemma [Dehn 1912] [Lyndon 1966] ‣ Consider a simple cycle in Σ̃ — no spurs, no self- intersections. (This is the interesting case.) ‣ Call a vertex convex if it is incident to exactly one interior corner. Let V + be the set of convex vertices.
Dehn’s lemma [Dehn 1912] [Lyndon 1966] ‣ Assign angle ¼ to every corner. ▹ κ ( v ) = ¼ for every convex vertex v ▹ κ ( v ) ≤ 0 for every non-convex vertex v ▹ κ ( f ) = 1– g < 0 for every face f ‣ Discrete Gauss-Bonnet ⇒ ∑ v κ ( v ) + ∑ f κ ( f ) = 1 ⇒ | V + | ≥ (4 g –4)| F | + 4. ⇒ Some face has ≥ 4 g –3 consecutive convex corners. □
Analysis ‣ Time for greedy reduction: ▹ Brute force: O( g 2 ) time per edge • At each step, compare last 2 g –2 edges to O( g ) patterns ▹ Smarter: O(1) amortized time per edge • Use a DFA for pattern matching • Charge O( g ) modification time to the 2 g –4 edges removed ‣ Total time for reduction is O( ℓ ʹ ) ‣ So overall time is O( gn + ℓ ʹ ) = O( gn + gℓ )
Linear time contractibility
System of quads [Lazarus Rivaud 2012] ‣ Add edges from center of P to corners, delete edges of P ‣ 2 vertices, 4 g edges, 2 g quadrilateral faces ‣ Edge in Σ → path of length ≤2 inside P ‣ Walk of length ℓ in Σ → walk of length ℓ ʹ ≤2 ℓ in quad system 1 2 b b a a a a a b 4 0 7 1 5 0 d c d c 7 3 c d 3 6 5 2 c c d d b b 4 6
System of quads [Lazarus Rivaud 2012] ‣ Add edges from center of P to corners, delete edges of P ‣ 2 vertices, 4 g edges, 2 g quadrilateral faces ‣ Edge in Σ → path of length ≤2 inside P ‣ Walk of length ℓ in Σ → walk of length ℓ ʹ ≤2 ℓ in quad system 1 2 b b a a a a a b 4 0 7 1 5 0 d c d c 7 3 c d 3 6 5 2 c c d d b b 4 6
Regular hyperbolic structure
Regular hyperbolic structure
Dehn’s algorithm ‣ Goal: Find a shortest cycle homotopic to the given cycle α. ‣ Algorithm: Greedily reduce α by local curve shortening. ‣ Hyperbolic structure ⇒ no local minima ! ‣ Thus α is contractible iff it reduces to nothing.
Turn sequence ‣ Consider a closed walk ( v 0 , e 1 , v 1 , e 2 , ..., v ℓ–1 , e ℓ ). ‣ Turn τ i = number of corners between e i –1 and e i in clockwise order around v i . 4 3 -3 2 -2 1 -1 0 ‣ Turn sequence (τ 0 , τ 1 , ..., τ ℓ ) is easy to compute in O( ℓ ) time.
Run-length encoding ‣ Record lengths of maximal runs of equal turns (1, 2 4 , 1 2 , 2) = (1, 2, 2, 2, 2 , 1, 1, 2) ‣ Easy to compute in O(ℓ) time ‣ The rest of the algorithm manipulates only run-length encoded turn sequences.
Spurs and brackets ‣ Left bracket: 1 2 k 1 ‣ Spur: 0 ‣ Right bracket: -1 -2 k -1 x x -1 1 2 -2 0 2 -2 x y 2 -2 2 -2 1 -1 y y
Bracket Lemma [Gersten Short 1990] ‣ Every nontrivial contractible cycle in Q̃ contains either a spur or a bracket. ‣ Corollary: A cycle is contractible iff it reduces to nothing. x x -1 1 2 -2 0 2 -2 x y 2 -2 2 -2 1 -1 y y
Bracket Lemma ‣ Consider a simple cycle in Q̃ — no spurs, no self- intersections. (This is the interesting case.) ‣ Label boundary vertices as follows: Convex: next to 1 corner Flat: next to 2 corners Concave: next to ≥3 corners
Bracket Lemma ‣ Consider a simple cycle in Q̃ — no spurs, no self- intersections. (This is the interesting case.) ‣ Label boundary vertices as follows: Convex: next to 1 corner Flat: next to 2 corners Concave: next to ≥3 corners
Bracket Lemma ‣ Consider a simple cycle in Q̃ — no spurs, no self- intersections. (This is the interesting case.) ‣ Label boundary vertices as follows: Convex: next to 1 corner Flat: next to 2 corners Concave: next to ≥3 corners
Bracket Lemma ‣ Assign angle ¼ to every corner. Faces are squares ! • Every face has curvature 0 Convex vertices have curvature +¼ Flat vertices have curvature 0 Concave vertices have curvature ≤ –¼ Interior vertices have curvature ≤ –1 ‣ Combinatorial Gauss-Bonnet: ∑ v κ ( v ) = 1 ⇒ #convex ≥ #concave + 4 ⇒ There are at least four brackets!
Bracket Lemma ‣ Assign angle ¼ to every corner. Faces are squares ! • Every face has curvature 0 Convex vertices have curvature +¼ Flat vertices have curvature 0 Concave vertices have curvature ≤ –¼ Interior vertices have curvature ≤ –1 ‣ Combinatorial Gauss-Bonnet: ∑ v κ ( v ) = 1 ⇒ #convex ≥ #concave + 4 ⇒ There are at least four brackets!
Bracket Lemma ‣ Assign angle ¼ to every corner. Faces are squares ! • Every face has curvature 0 Convex vertices have curvature +¼ Flat vertices have curvature 0 Concave vertices have curvature ≤ –¼ Interior vertices have curvature ≤ –1 ‣ Combinatorial Gauss-Bonnet: ∑ v κ ( v ) = 1 ⇒ #convex ≥ #concave + 4 ⇒ There are at least four brackets!
Elementary reductions ‣ Left bracket: x 1 2 k 1 y → x –1 -2 k y –1 ‣ Spur: x 0 y → x + y ‣ Right bracket: x -1 -2 k -1 y → x +1 2 k y +1 x x − 1 0 x x+1 -1 1 x y 2 -2 -2 2 2 -2 -2 2 2 -2 -2 2 x+y 2 -2 -2 2 1 -1 y y y − 1 y+1
Cyclic elementary reductions ‣ Left bracket: ( x 1 2 k 1) → ( x –2 -2 k ) or (1 2 k ) → (-3 -2 k–2 ) ‣ Double spur: (0 0) → ( ) ‣ Right bracket: ( x -1 -2 k -1) → ( x +1 2 k ) or (-1 -2 k ) → (3 2 k–2 ) 2 1 -1 -2 2 -2 x − 2 x+2 x x 1 -1 -2 2 -2 2 -2 -2 2 -2 2 2 2 -2 -2 -2 -2 2 2 2 0 0 2 2 -2 -2 2 -2 2 2 1 -1 -2 -2 2 -2 2 -2 -2 -3 3 2 -2 2 -2 -2 2 -2 2 2 2 -2 -2 -2 -2 2 2 2 2 -2 2 -2 2 -2
Reduction algorithm mark all runs dirty i ← 0 repeat if runs i –4 .. i contain a spur or bracket perform an elementary reduction mark the modified runs dirty i ← max{0, i –5} else mark run i clean i ← i +1 until all runs are marked clean
Reduction takes O(ℓ) time ‣ Turn sequence is run-length encoded ⇒ Each iteration takes O(1) time ‣ Each elementary reduction decreases length by 2 ⇒ At most ℓ/2 elementary reductions ‣ Each elementary reduction creates or modifies ≤5 runs ⇒ At most ℓ + 5ℓ/2 runs to process ⇒ At most 4ℓ iterations
Free homotopy
The homotopy problem ‣ Given two paths or cycles α and β in an orientable surface, can α be continuously deformed into β? No Yes ‣ Are α and β freely homotopic / in the same free homotopy class?
Free homotopy ‣ Goal: Transform each cycle into the unique canonical cycle in its homotopy class. Two cycles are homotopic if and only if they yield the same canonical cycle. ▹ With a smooth hyperbolic metric, each homotopy class has a unique shortest cycle. [Dehn 1911-12] ▹ But in our discrete hyperbolic metric, shortest cycles are not unique. ‣ Algorithm: After shrinking the cycles, shift them to the right as far as possible without increasing their length. [Lazarus Rivaud 2012]
[Lazarus Rivaud 2012] Canonical cycles [Erickson Whittlesey 2013] ‣ A reduced cycle is canonical iff (1) it has no turn –1 (2) not all turns are –2 ‣ Canonical cycles are as far to the right as possible. -2 -2 -2 -1 y y +1 2 2 2 2 2 -2 -2 x 2 2 2 x +1 1 -2 -2 -2
Elementary right shifts ‣ x -2 s -1 -2 t y → x +1 1 2 s −1 3 2 t −1 1 y +1 (if s >0, t >0) ‣ x -1 -2 t y → x +1 2 t 1 y +1 ‣ x -2 s -1 y → x +1 1 2 s y +1 -1 -2 -2 -2 -2 y -2 -1 y y+1 -1 -2 -2 -2 -2 y -2 -2 x y+1 -2 2 -2 x 2 y+1 -2 3 2 2 2 1 2 2 x x+1 x+1 1 2 2 2 2 1 2 x+1 1
Cyclic elementary right shifts ‣ ( x -2 s -1 -2 t ) → ( x +2 1 2 s −1 3 2 t −1 1) (s>0, t>0, x≠-3) ‣ ( x -1 -2 t ) → ( x +2 2 t 1) (x≠-3) ‣ ( x -2 s -1 ) → ( x +2 1 2 s ) (x≠-3) ‣ (-3 -2 s -1 -2 t ) → (1 2 s 3 2 t ) ‣ (-2 t ) → (2 t ) -2 -2 -2 -2 -2 -2 -2 -2 x -2 -2 -2 -3 -2 -2 -2 -2 -2 -2 -2 x -2 -1 -2 -2 -2 -2 -2 -2 -2 -1 x -2 -2 -2 -2 -2 -2 -2 -2 -2 -2 -1 -2 -2 -2 -2 -2 -2 -2 -1 -2 -2 -2 -2 -2 x+2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2 2 2 2 x+2 2 2 1 2 2 2 2 2 2 2 2 2 2 1 1 x+2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 3 2 2 2
Homotopy algorithm ‣ Reduce both cycles: O( ℓ ) time ‣ Make reduced cycles canonical via right shifts: O(ℓ) time ▹ Run length encoding ⇒ each shift takes O(1) time ▹ No backtracking required ‣ Cycles are homotopic iff same canonical cycle: O(ℓ) time
[Lazarus Rivaud 2012] Correctness [Erickson Whittlesey 2013] ‣ Lemma: Each free homotopy class contains exactly one canonical cycle. ‣ Corollary: Two cycles are freely homotopic iff they yield the same canonical cycle. ‣ Proof uses Combinatorial Gauss-Bonnet Theorem again -1 -2 -2 -2 R S S S Δ x -1 -2 -2 -2 -2 -3 Δ x x R S S S S L 3 2 2 -2 A' A' y 3 2 2 2 2 1 S -2 2 S -2 -3 2 S S L 2 2 1
Shortest homotopic paths
Shortest homotopic curves ‣ Given an arbitrary surface map Σ with weighted edges and a walk α in Σ, find the shortest walk homotopic to α. ▹ Algorithms we’ve just seen solve this problem when Σ is an unweighted system of loops or system of quads
Tight hexagonal decomposition 4 g cycles, each as short as possible in its homotopy class, that decompose Σ into “right-angled hexagons” [Colin de Verdière, Erickson 2006]
Universal cover The hexagonal decomposition lifts to a regular tiling of the hyperbolic plane with right- angled hexagons. [Colin de Verdière, Erickson 2006]
Universal cover The hexagonal decomposition lifts to a regular tiling of the hyperbolic plane with right- angled hexagons. M. C. Escher, Circle Limit IV: Heaven and Hell (1960)
Universal cover The hexagonal decomposition lifts to a regular tiling of the hyperbolic plane with right- angled hexagons. M. C. Escher, Circle Limit IV: Heaven and Hell (1960)
Universal cover ‣ Each cycle in the tight hexagonal decomposition lifts to a line —an infinite shortest path—in Σ̃. ‣ Shortest paths in Σ̃ cross each line at most once. [Colin de Verdière, Erickson 2006]
Universal cover ‣ First encode the input path by the sequence of lines that it crosses. ‣ Then reduce the crossing sequence using small cancellation. ‣ The reduced crossing sequence lists the lines that π̃ crosses an odd number of times. [Colin de Verdière, Erickson 2006]
Universal cover ‣ First encode the input path by the sequence of lines that it crosses. ‣ Then reduce the crossing sequence using small cancellation. ‣ The reduced crossing sequence lists the lines that π̃ crosses an odd number of times. [Colin de Verdière, Erickson 2006]
Universal cover ‣ First encode the input path by the sequence of lines that it crosses. ‣ Then reduce the crossing sequence using small cancellation. ‣ The reduced crossing sequence lists the lines that π̃ crosses an odd number of times. [Colin de Verdière, Erickson 2006]
Relevant region ‣ Hexagons in universal cover containing all reduced paths between endpoints of π̃. ‣ Convex hyperbolic polygon ‣ Discrete Gauss-Bonnet ⇒ O( x ) relevant hexagons ‣ Comput from the reduced crossing sequence in O(x) time. [Colin de Verdière, Erickson 2006]
Relevant region ‣ Hexagons in universal cover containing all reduced paths between endpoints of π̃. ‣ Convex hyperbolic polygon ‣ Discrete Gauss-Bonnet ⇒ O( x ) relevant hexagons ‣ Comput from the reduced crossing sequence in O(x) time. [Colin de Verdière, Erickson 2006]
Relevant region ‣ Hexagons in universal cover containing all reduced paths between endpoints of π̃. ‣ Convex hyperbolic polygon ‣ Discrete Gauss-Bonnet ⇒ O( x ) relevant hexagons ‣ Comput from the reduced crossing sequence in O(x) time. [Colin de Verdière, Erickson 2006]
Shortest homotopic paths ‣ Given a path π with ℓ edges in a surface map, the shortest path homotopic to π can be found in O( gn log n + gnℓ ) time. ▹ Build a tight hexagonal decomposition — O( gn log n ) time ▹ Compute the crossing sequence of π — O( x ) = O( gℓ ) time ▹ Reduce crossing sequence via small cancellation — O( x ) time ▹ Build relevant region of Σ̃ — O( xn ) time ▹ Find shortest path in relevant region — O( xn ) time [Colin de Verdière, Erickson 2006]
Higher dimensions?
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